Problem: Solve for $x$: $\left(\frac{1}{4}\right)^{2x+8} = (16)^{2x+5}$.
We begin by writing $\frac{1}{4}$ and $16$ as powers of $2$. \begin{align*}
(2^{-2})^{2x+8} & = (2^4)^{2x+5} \\
2^{-4x-16} & = 2^{8x + 20} \\
-4x - 16 & = 8x + 20 \\
x & = \boxed{-3}
\end{align*}